answer：Alright, so I've got this algebra problem here, and I need to find all integers a, b, and c such that the equation [(x - a)(x - 11) + 2 = (x + b)(x + c)] holds true. Hmm, okay, let me try to unpack this step by step.First off, I know that both sides of the equation are quadratic expressions in terms of x. So, if two quadratics are equal for all x, their corresponding coefficients must be equal. That means the coefficients of x², x, and the constant term must be the same on both sides. That's a key point.Let me write out both sides in expanded form to compare them. Starting with the left side: [(x - a)(x - 11) + 2]. I'll expand this first. Multiplying (x - a) and (x - 11), I get:[x cdot x + x cdot (-11) + (-a) cdot x + (-a) cdot (-11)]Which simplifies to:[x^2 - 11x - a x + 11a]Combine like terms:[x^2 - (11 + a)x + 11a]Then, adding the +2 from the original equation:[x^2 - (11 + a)x + 11a + 2]Okay, so the left side is (x^2 - (11 + a)x + (11a + 2)).Now, the right side is [(x + b)(x + c)]. Let me expand that as well:[x cdot x + x cdot c + b cdot x + b cdot c]Which simplifies to:[x^2 + (b + c)x + bc]So, the right side is (x^2 + (b + c)x + bc).Since these two quadratics are equal for all x, their coefficients must be equal. That gives us the following system of equations:1. Coefficient of x²: 1 = 1. Well, that's already satisfied.2. Coefficient of x: -(11 + a) = b + c3. Constant term: 11a + 2 = bcSo, now we have two equations:1. -(11 + a) = b + c2. 11a + 2 = bcOur goal is to find integers a, b, and c that satisfy both equations.Let me write these equations again for clarity:1. b + c = - (11 + a) --> Equation (1)2. bc = 11a + 2 --> Equation (2)So, we have two equations with three variables, but since we're looking for integer solutions, maybe we can find possible integer pairs (b, c) that satisfy Equation (2) and then see if they also satisfy Equation (1).Alternatively, perhaps we can express a in terms of b and c from Equation (1) and substitute into Equation (2). Let me try that.From Equation (1):b + c = - (11 + a) => a = - (b + c + 11) --> Equation (1a)Now, substitute a into Equation (2):bc = 11a + 2 => bc = 11*(- (b + c + 11)) + 2 => bc = -11(b + c + 11) + 2 => bc = -11b -11c - 121 + 2 => bc = -11b -11c - 119Let me bring all terms to one side:bc + 11b + 11c + 119 = 0Hmm, this looks a bit complicated, but maybe we can factor this expression. Let me see if I can factor by grouping.Looking at bc + 11b + 11c + 119, perhaps group the first two terms and the last two terms:(bc + 11b) + (11c + 119) = b(c + 11) + 11(c + 11) = (b + 11)(c + 11)Ah! So, bc + 11b + 11c + 119 factors into (b + 11)(c + 11). Therefore, the equation becomes:(b + 11)(c + 11) = 0Wait, but that's not correct because when I factor, I should have:(b + 11)(c + 11) = bc + 11b + 11c + 121But in our case, we have bc + 11b + 11c + 119, which is 2 less than 121. So, actually:(b + 11)(c + 11) = 121 But bc + 11b + 11c + 119 = (b + 11)(c + 11) - 2 = 0 Wait, that might not be the right approach.Wait, let's go back.We had:bc + 11b + 11c + 119 = 0But (b + 11)(c + 11) = bc + 11b + 11c + 121So, if we subtract 2 from both sides:(b + 11)(c + 11) - 2 = bc + 11b + 11c + 119Which is equal to 0.Therefore:(b + 11)(c + 11) - 2 = 0 => (b + 11)(c + 11) = 2Ah, that's better. So, (b + 11)(c + 11) = 2.Now, since b and c are integers, (b + 11) and (c + 11) must be integer factors of 2.So, the possible factor pairs of 2 (considering both positive and negative factors) are:1. (1, 2)2. (2, 1)3. (-1, -2)4. (-2, -1)So, these are the possible pairs for (b + 11, c + 11). Let's consider each case.Case 1: b + 11 = 1 and c + 11 = 2 Then, b = 1 - 11 = -10 c = 2 - 11 = -9 So, b = -10, c = -9Case 2: b + 11 = 2 and c + 11 = 1 Then, b = 2 - 11 = -9 c = 1 - 11 = -10 So, b = -9, c = -10Case 3: b + 11 = -1 and c + 11 = -2 Then, b = -1 - 11 = -12 c = -2 - 11 = -13 So, b = -12, c = -13Case 4: b + 11 = -2 and c + 11 = -1 Then, b = -2 - 11 = -13 c = -1 - 11 = -12 So, b = -13, c = -12So, we have four possible pairs for (b, c):1. (-10, -9)2. (-9, -10)3. (-12, -13)4. (-13, -12)Now, for each of these, we can find a using Equation (1a):a = - (b + c + 11)Let's compute a for each case.Case 1: b = -10, c = -9 a = - ((-10) + (-9) + 11) = - (-19 + 11) = - (-8) = 8Case 2: b = -9, c = -10 a = - ((-9) + (-10) + 11) = - (-19 + 11) = - (-8) = 8Case 3: b = -12, c = -13 a = - ((-12) + (-13) + 11) = - (-25 + 11) = - (-14) = 14Case 4: b = -13, c = -12 a = - ((-13) + (-12) + 11) = - (-25 + 11) = - (-14) = 14So, for each pair of (b, c), we get a corresponding a:- Cases 1 and 2 give a = 8- Cases 3 and 4 give a = 14Now, let's check if these solutions satisfy the original equation.First, let's check a = 8, b = -10, c = -9.Left side: (x - 8)(x - 11) + 2 Let's compute (x - 8)(x - 11): = x² - 11x - 8x + 88 = x² - 19x + 88 Add 2: x² - 19x + 90Right side: (x - 10)(x - 9) = x² - 9x -10x + 90 = x² - 19x + 90They match. So, a = 8, b = -10, c = -9 is a valid solution.Similarly, for a = 8, b = -9, c = -10, the right side is (x - 9)(x - 10), which is the same as (x -10)(x -9), so it's the same quadratic, so it's also valid.Now, let's check a =14, b = -12, c = -13.Left side: (x -14)(x -11) + 2 Compute (x -14)(x -11): = x² -11x -14x + 154 = x² -25x + 154 Add 2: x² -25x + 156Right side: (x -12)(x -13) = x² -13x -12x + 156 = x² -25x + 156They match. So, a =14, b = -12, c = -13 is a valid solution.Similarly, for a =14, b = -13, c = -12, the right side is (x -13)(x -12), which is the same as (x -12)(x -13), so it's also valid.So, we have two sets of solutions:1. a = 8, b = -10, c = -92. a = 14, b = -12, c = -13Wait, but the problem asks for all integers a, b, and c. So, both sets are valid.But hold on, in the initial problem statement, when I set x = 11, I got (b +11)(c +11) = 2, which led me to four possible pairs for (b, c). Then, solving for a, I got a =8 and a=14.But in the initial attempt, the user only found a=13, which seems incorrect because in my calculation, a=8 and a=14 are the solutions.Wait, maybe I made a mistake in the initial steps. Let me double-check.Wait, in the initial problem, the user set x=11 and got (b +11)(c +11)=2, which is correct. Then, they considered factor pairs (2,1) and (-2,-1), leading to b=-9, c=-10 and b=-13, c=-12.But when they plugged in b=-9, c=-10, they tried to equate the quadratics and found a non-integer a, which they discarded. Similarly, when they tried b=-13, c=-12, they found a=13.But in my approach, I considered all four factor pairs, including (1,2) and (-1,-2), which gave me a=8 and a=14.Wait, perhaps the user missed some factor pairs? Because 2 can also be factored as (1,2) and (-1,-2), which would give different b and c values, leading to a=8 and a=14.So, in the initial attempt, the user only considered (2,1) and (-2,-1), but missed (1,2) and (-1,-2), which are also valid factor pairs.Therefore, the complete solution should include both a=8 and a=14, along with their corresponding b and c values.So, to summarize, the integer solutions are:1. a=8, b=-10, c=-92. a=8, b=-9, c=-103. a=14, b=-12, c=-134. a=14, b=-13, c=-12But since b and c are interchangeable in the right side, (x + b)(x + c) is the same as (x + c)(x + b), so the solutions with a=8 and a=14 are distinct, but the pairs (b,c) are just swapped.Therefore, the distinct solutions are a=8 and a=14, each with their respective b and c.But wait, let me check if a=8 and a=14 are the only solutions.Given that (b +11)(c +11)=2, and 2 has only four integer factor pairs: (1,2), (2,1), (-1,-2), (-2,-1). So, we've considered all possible factor pairs, leading to a=8 and a=14.Therefore, the integer solutions are a=8 and a=14, with corresponding b and c as above.But in the initial problem, the user only found a=13, which seems to be incorrect because in their calculation, when they set b=-13 and c=-12, they found a=13, but in reality, a should be 14.Wait, let me check their calculation.They had:When b=-13 and c=-12,Left side: (x - a)(x -11) +2 Right side: (x -13)(x -12) = x² -25x +156They set (x -11)^2 +2 = x² -22x +123, which is correct.Then, equating to x² -25x +156,They wrote:-22x +123 = -25x +156Then, solving:3x = 33 x=11Then, they concluded that for (x -a) term to hold, a=13.Wait, that seems incorrect because when x=11, the equation holds, but that doesn't necessarily determine a. Instead, we should equate the coefficients of x and the constant term.Wait, let's see.If we have:Left side: x² - (11 + a)x + (11a +2)Right side: x² -25x +156Therefore, equate coefficients:- (11 + a) = -25 => 11 + a =25 => a=14Similarly, constant term:11a +2 =156 =>11a=154 =>a=14So, a=14, which is consistent.But in the initial attempt, the user set x=11 and got (b +11)(c +11)=2, then tried to plug in b=-13, c=-12, and then equated the quadratics, but instead of equating coefficients, they set x=11 again, leading to x=11, which didn't help in finding a.Therefore, their approach was flawed because they didn't equate the coefficients properly, leading them to an incorrect a=13 instead of a=14.So, in conclusion, the correct integer solutions are a=8 and a=14, with corresponding b and c values.Therefore, the possible values of a are 8 and 14.But wait, the problem says "determine all integers a, b, and c". So, we need to list all possible triples (a,b,c).From our earlier analysis, the solutions are:1. a=8, b=-10, c=-92. a=8, b=-9, c=-103. a=14, b=-12, c=-134. a=14, b=-13, c=-12But since b and c are symmetric in the equation (x + b)(x + c), the pairs (b,c) are interchangeable. Therefore, the distinct solutions are two: one with a=8 and another with a=14.Hence, the integer solutions are:- a=8, b=-10, c=-9- a=14, b=-12, c=-13These are the only integer solutions that satisfy the given equation.To ensure there are no other solutions, we've considered all factor pairs of 2, both positive and negative, and found that only these two values of a satisfy the conditions. Therefore, we can confidently say that these are the only solutions.In summary, the integers a, b, and c that satisfy the equation are:- a=8, b=-10, c=-9- a=14, b=-12, c=-13These are the complete set of solutions.

answer：First, I need to determine the individual costs of a pen and a pencil based on the given ratio and total cost.Let's assume the cost of one pencil is ( x ) rupees. According to the ratio of 5:1, the cost of one pen would then be ( 5x ) rupees.Next, I'll set up an equation based on the total cost of 3 pens and 5 pencils:[3 times 5x + 5 times x = 100]Simplifying this equation:[15x + 5x = 100 20x = 100 x = 5]So, the cost of one pencil is Rs. 5, and the cost of one pen is:[5x = 5 times 5 = 25 text{ rupees}]Finally, to find the cost of one dozen pens (12 pens):[12 times 25 = 300 text{ rupees}]

answer：The library automated return system starts processing returns at 9:30 AM. By 12:00 PM, it has processed half of the day's returns. To determine when it will finish processing all the returns, we can follow these steps:1. **Calculate the time taken to process half of the returns:** - The system starts at 9:30 AM and processes half of the returns by 12:00 PM. - The duration from 9:30 AM to 12:00 PM is 2 hours and 30 minutes, which is equivalent to 2.5 hours.2. **Calculate the total time required to process all returns:** - If 2.5 hours is needed for half of the returns, then the total time for all returns is: [ text{Total time} = 2 times 2.5 text{ hours} = 5 text{ hours} ]3. **Determine the completion time:** - Adding 5 hours to the start time of 9:30 AM: [ text{Completion time} = text{9:30 AM} + 5 text{ hours} = text{2:30 PM} ]Thus, the library automated return system will finish processing all the returns by 2:30 PM.boxed{mathrm{(B)} text{2:30} {smalltext{PM}}}

answer：Alright, so I have this problem about a parabola, points A and B on it, and their perpendicular bisector intersecting the x-axis at point D(4,0). I need to find the maximum value of the distance between A and B, which is |AB|. Hmm, okay, let me try to break this down.First, I know that the equation of the parabola is y² = 4x. That's a standard right-opening parabola with vertex at the origin. The general form of such a parabola is y² = 4ax, where 'a' is the distance from the vertex to the focus. In this case, 4a = 4, so a = 1. That means the focus is at (1,0), and the directrix is the line x = -1.Now, points A and B lie on this parabola. Let me denote A as (x₁, y₁) and B as (x₂, y₂). Since both points are on the parabola, they satisfy the equation y₁² = 4x₁ and y₂² = 4x₂.The problem mentions the perpendicular bisector of segment AB intersects the x-axis at D(4,0). So, I need to find the perpendicular bisector of AB and set its intersection with the x-axis to be (4,0). From this condition, I should be able to derive some relationship between y₁ and y₂, which will help me find the maximum distance between A and B.Let me recall that the perpendicular bisector of a segment passes through the midpoint of the segment and has a slope that is the negative reciprocal of the segment's slope. So, first, I need to find the midpoint M of AB.The midpoint M will have coordinates ((x₁ + x₂)/2, (y₁ + y₂)/2). Let me denote this as (x₀, y₀). So, x₀ = (x₁ + x₂)/2 and y₀ = (y₁ + y₂)/2.Next, I need the slope of AB. The slope k of AB is (y₂ - y₁)/(x₂ - x₁). Since both points are on the parabola, I can express x₁ and x₂ in terms of y₁ and y₂: x₁ = y₁²/4 and x₂ = y₂²/4. So, substituting these into the slope formula:k = (y₂ - y₁) / (y₂²/4 - y₁²/4) = (y₂ - y₁) / [(y₂² - y₁²)/4] = 4(y₂ - y₁)/(y₂² - y₁²)I can factor the denominator as (y₂ - y₁)(y₂ + y₁), so:k = 4(y₂ - y₁)/[(y₂ - y₁)(y₂ + y₁)] = 4/(y₂ + y₁)So, the slope of AB is 4/(y₁ + y₂). Therefore, the slope of the perpendicular bisector is the negative reciprocal, which is -(y₁ + y₂)/4.Now, the perpendicular bisector passes through the midpoint M(x₀, y₀). So, the equation of the perpendicular bisector is:(y - y₀) = [-(y₁ + y₂)/4](x - x₀)We know that this line intersects the x-axis at D(4,0). So, when y = 0, x = 4. Plugging these into the equation:0 - y₀ = [-(y₁ + y₂)/4](4 - x₀)Simplify:-y₀ = [-(y₁ + y₂)/4](4 - x₀)Multiply both sides by 4 to eliminate the denominator:-4y₀ = -(y₁ + y₂)(4 - x₀)Simplify the negatives:4y₀ = (y₁ + y₂)(4 - x₀)But y₀ = (y₁ + y₂)/2, so substitute that in:4*(y₁ + y₂)/2 = (y₁ + y₂)(4 - x₀)Simplify left side:2(y₁ + y₂) = (y₁ + y₂)(4 - x₀)Assuming y₁ + y₂ ≠ 0 (since if y₁ + y₂ = 0, then the slope would be undefined, which might not be the case here), we can divide both sides by (y₁ + y₂):2 = 4 - x₀So, x₀ = 4 - 2 = 2Therefore, the x-coordinate of the midpoint M is 2. So, x₀ = 2.But x₀ is the average of x₁ and x₂, so:(x₁ + x₂)/2 = 2 => x₁ + x₂ = 4So, the sum of the x-coordinates of points A and B is 4.Now, since x₁ = y₁²/4 and x₂ = y₂²/4, we have:y₁²/4 + y₂²/4 = 4 => y₁² + y₂² = 16So, the sum of the squares of y₁ and y₂ is 16.Now, I need to find the maximum value of |AB|. The distance between A and B is:|AB| = sqrt[(x₂ - x₁)² + (y₂ - y₁)²]Since x₁ = y₁²/4 and x₂ = y₂²/4, substitute these in:|AB| = sqrt[(y₂²/4 - y₁²/4)² + (y₂ - y₁)²] = sqrt[((y₂² - y₁²)/4)² + (y₂ - y₁)²]Let me factor out (y₂ - y₁)² from both terms inside the square root:= sqrt[(y₂ - y₁)² * [( (y₂ + y₁)/4 )² + 1]]So, |AB| = |y₂ - y₁| * sqrt[ ( (y₂ + y₁)/4 )² + 1 ]Let me denote S = y₁ + y₂ and D = y₂ - y₁.We know from earlier that y₁² + y₂² = 16.Also, S² = (y₁ + y₂)² = y₁² + 2y₁y₂ + y₂² = 16 + 2y₁y₂Similarly, D² = (y₂ - y₁)² = y₂² - 2y₁y₂ + y₁² = 16 - 2y₁y₂So, S² + D² = 32But I'm not sure if that helps directly. Let me think.We have |AB| = |D| * sqrt[ (S/4)² + 1 ]So, |AB| = |D| * sqrt( S²/16 + 1 )But S² = 16 + 2y₁y₂, so:|AB| = |D| * sqrt( (16 + 2y₁y₂)/16 + 1 ) = |D| * sqrt(1 + (2y₁y₂)/16 + 1 )Wait, that seems messy. Maybe another approach.Alternatively, since we have y₁² + y₂² = 16, and we need to maximize |AB|, which is |D| * sqrt( (S/4)² + 1 )Let me express everything in terms of S and D.We have:y₁² + y₂² = 16Also, S = y₁ + y₂D = y₂ - y₁So, y₁ = (S - D)/2y₂ = (S + D)/2Then, y₁² + y₂² = [(S - D)/2]^2 + [(S + D)/2]^2 = [ (S² - 2SD + D²) + (S² + 2SD + D²) ] / 4 = [2S² + 2D²]/4 = (S² + D²)/2 = 16So, (S² + D²)/2 = 16 => S² + D² = 32So, we have S² + D² = 32Now, |AB| = |D| * sqrt( (S/4)^2 + 1 ) = |D| * sqrt( S²/16 + 1 )Let me denote |AB| = |D| * sqrt( S²/16 + 1 )We need to maximize this expression given that S² + D² = 32.Let me express S² in terms of D²: S² = 32 - D²So, substitute into |AB|:|AB| = |D| * sqrt( (32 - D²)/16 + 1 ) = |D| * sqrt( 2 - D²/16 + 1 ) = |D| * sqrt( 3 - D²/16 )Wait, that doesn't seem right. Let me check the substitution:sqrt( S²/16 + 1 ) = sqrt( (32 - D²)/16 + 1 ) = sqrt( 32/16 - D²/16 + 1 ) = sqrt( 2 - D²/16 + 1 ) = sqrt(3 - D²/16 )Yes, that's correct.So, |AB| = |D| * sqrt(3 - D²/16 )Since |D| is positive, we can write |AB| = D * sqrt(3 - D²/16 ), where D > 0.Now, to maximize |AB|, we can consider D as a variable and find its maximum.Let me set f(D) = D * sqrt(3 - D²/16 )To find the maximum of f(D), we can take the derivative and set it to zero.But since this is a bit involved, maybe we can square the function to make it easier.Let me define f(D)^2 = D² * (3 - D²/16 ) = 3D² - D⁴/16Let me denote g(D) = 3D² - D⁴/16To find the maximum of g(D), take derivative:g'(D) = 6D - (4D³)/16 = 6D - D³/4Set g'(D) = 0:6D - D³/4 = 0 => D(6 - D²/4) = 0So, D = 0 or 6 - D²/4 = 0 => D² = 24 => D = sqrt(24) = 2*sqrt(6)But D = 0 would give |AB| = 0, which is the minimum, so the maximum occurs at D = 2*sqrt(6)Now, check if this is valid. Since S² + D² = 32, and D² = 24, then S² = 8, so S = ±2*sqrt(2)So, S = y₁ + y₂ = ±2*sqrt(2)Now, let's compute |AB|:|AB| = D * sqrt(3 - D²/16 ) = 2*sqrt(6) * sqrt(3 - 24/16 ) = 2*sqrt(6) * sqrt(3 - 1.5 ) = 2*sqrt(6) * sqrt(1.5 )Simplify sqrt(1.5 ) = sqrt(3/2 ) = sqrt(3)/sqrt(2 )So, |AB| = 2*sqrt(6) * sqrt(3)/sqrt(2 ) = 2*sqrt(6)*sqrt(3)/sqrt(2 )Simplify sqrt(6)/sqrt(2 ) = sqrt(3 )So, |AB| = 2*sqrt(3)*sqrt(3 ) = 2*3 = 6Wait, that's interesting. So, the maximum |AB| is 6.But let me verify this because sometimes when we square functions, we might introduce extraneous solutions, but in this case, since we're dealing with distances, it should be fine.Alternatively, another approach is to consider that the maximum distance between two points on a parabola with a given condition often relates to the focal length or other geometric properties.Given that the focus is at (1,0), and the directrix is x = -1, perhaps there's a relationship involving the focus.But in this case, since the perpendicular bisector intersects the x-axis at (4,0), which is to the right of the focus, it might not directly relate to the focal length.But regardless, through the algebraic approach, I arrived at |AB| = 6 as the maximum distance.Let me just recap:1. Expressed points A and B on the parabola in terms of y₁ and y₂.2. Found the midpoint M and the slope of AB.3. Derived the equation of the perpendicular bisector and used the condition that it passes through D(4,0) to find that x₀ = 2.4. From x₀ = 2, deduced that x₁ + x₂ = 4, leading to y₁² + y₂² = 16.5. Expressed |AB| in terms of y₁ and y₂, then converted to variables S and D.6. Used the constraint S² + D² = 32 to express |AB| solely in terms of D.7. Maximized |AB| by taking the derivative and found that the maximum occurs at D = 2*sqrt(6), leading to |AB| = 6.Therefore, the maximum value of |AB| is 6.

answer：Alright, so I have this problem here: I need to prove that if the sum of (cos x_j + 1) for j from 1 to n is an odd integer, then the sum of sin x_j from 1 to n is at least 1. All the x_j are between 0 and pi. Hmm, okay, let's break this down.First, let's understand what the sum of (cos x_j + 1) being an odd integer means. Each term in the sum is (cos x_j + 1). Since cos x_j ranges from -1 to 1, adding 1 to it makes each term range from 0 to 2. So, each term is between 0 and 2. The sum of these terms is an odd integer. That means the total sum is something like 1, 3, 5, etc.Now, I need to show that the sum of sin x_j is at least 1. So, if I can find a relationship between the sum of (cos x_j + 1) and the sum of sin x_j, maybe I can establish that the latter has to be at least 1.Let me think about individual terms. For each x_j, cos x_j + 1 can be 0, 1, or 2. But since the total sum is an odd integer, the sum must be something like 1, 3, 5, etc. So, maybe each term contributes either 0, 1, or 2, but the total has to be odd.Wait, but cos x_j + 1 is always non-negative because cos x_j is between -1 and 1, so adding 1 makes it between 0 and 2. So, each term is non-negative, and the total sum is an odd integer.Now, if the sum is an odd integer, that means the number of terms that contribute 1 or 2 must result in an odd total. Let me think about how that can happen.If all the terms were 1, the sum would be n, which is an integer. But it's an odd integer, so n must be odd? Wait, no, because each term could be either 0, 1, or 2. So, it's not necessarily that n is odd, but the total sum is odd.For example, if n=2, and one term is 2 and the other is 1, the total sum is 3, which is odd. Similarly, if n=3, and all three terms are 1, the sum is 3, which is odd. So, it's not just about the number of terms, but how each term contributes.Now, I need to relate this to the sum of sin x_j. Let's recall that for x in [0, pi], sin x is non-negative, so the sum of sin x_j will also be non-negative.I wonder if I can use some inequality or identity that relates cos x and sin x. For example, sin^2 x + cos^2 x = 1. Maybe that can help.But I'm dealing with sums, not products or squares. Hmm.Alternatively, maybe I can consider the possible values of cos x_j + 1 and sin x_j for x_j in [0, pi].Let's list possible values:- If cos x_j + 1 = 0, then cos x_j = -1, which implies x_j = pi. Then sin x_j = 0.- If cos x_j + 1 = 1, then cos x_j = 0, which implies x_j = pi/2. Then sin x_j = 1.- If cos x_j + 1 = 2, then cos x_j = 1, which implies x_j = 0. Then sin x_j = 0.Wait a second, so each term in the sum (cos x_j + 1) can be 0, 1, or 2, corresponding to x_j being pi, pi/2, or 0, respectively. And in those cases, sin x_j is 0, 1, or 0.So, if I have a sum of (cos x_j + 1) being an odd integer, that means the number of terms where cos x_j + 1 = 1 must be odd because each such term contributes 1, and the others contribute 0 or 2, which are even.Wait, let's see:Each term is either 0, 1, or 2.If I have k terms where cos x_j + 1 = 1, then those contribute k to the sum.The other terms contribute either 0 or 2, which are even numbers.So, the total sum is k + (sum of even numbers). Since the total sum is odd, k must be odd because even + odd = odd.Therefore, the number of terms where cos x_j + 1 = 1 must be odd.But each such term corresponds to x_j = pi/2, and sin x_j = 1.So, if there are k such terms, each contributing 1 to the sum of sin x_j, then the total sum of sin x_j is at least k.But k is odd, so the smallest odd number is 1. Therefore, the sum of sin x_j is at least 1.Wait, is that it? Let me check.If the sum of (cos x_j + 1) is odd, then the number of terms where cos x_j + 1 = 1 must be odd. Each such term contributes 1 to the sum of sin x_j. Therefore, the total sum of sin x_j is at least 1.Yes, that seems to make sense.But let me think about it more carefully. Suppose n=1. Then, the sum is (cos x1 +1). For it to be odd, it must be 1, because cos x1 +1 can be 0,1,2. So, cos x1 +1=1 implies cos x1=0, so x1=pi/2, and sin x1=1. So, sum of sin x_j=1, which is >=1.If n=2, and the sum is 3, which is odd. How can that happen? Each term is 0,1, or 2. So, possible combinations:- One term is 2 and one term is 1: sum=3.- Or three terms of 1, but n=2, so that's not possible.Wait, n=2, sum=3: one term is 2, one term is1.So, one x_j=0 (sin x_j=0), and one x_j=pi/2 (sin x_j=1). So, sum of sin x_j=1.Alternatively, if n=3, sum=3: three terms of 1, so three x_j=pi/2, sum of sin x_j=3.Or, one term=2, one term=1, and one term=0: sum=3. Then, sum of sin x_j=1.So, in all cases, the sum of sin x_j is at least 1.Wait, but what if n=3, and two terms are 2, and one term is -1? Wait, no, cos x_j +1 can't be negative because cos x_j >=-1, so cos x_j +1 >=0.So, all terms are non-negative. So, the sum is non-negative.Therefore, the sum being odd implies that the number of terms where cos x_j +1=1 is odd, each contributing 1 to the sum of sin x_j, so the total sum is at least 1.Therefore, the conclusion holds.I think that's the gist of it. So, the key idea is that the sum being odd implies an odd number of terms where cos x_j +1=1, each contributing 1 to the sum of sin x_j, hence the total sum is at least 1.

answer：Alright, so I have this problem that says if there are ( n ) points not all on the same line, then there are at least ( n ) different lines connecting them. Hmm, okay, let me try to wrap my head around this.First, I need to understand what it means for points not to be on the same line. That means no single straight line can pass through all ( n ) points. So, they're scattered in some way, maybe forming a shape like a triangle or something more complex.Now, the problem is asking about the number of different lines connecting these points. I know that if you have two points, there's exactly one line connecting them. So, for ( n ) points, the total number of possible lines would be ( binom{n}{2} ), which is ( frac{n(n-1)}{2} ). But that's the maximum number of lines if no three points are on the same line. However, the problem is saying that even if some points are on the same line, as long as not all ( n ) are on the same line, there are at least ( n ) different lines.Wait, so if some points are on the same line, that would reduce the total number of lines, right? Because multiple points on the same line would mean that line is shared by multiple pairs of points. But the problem is saying that even with some overlapping lines, the total number of distinct lines is still at least ( n ).Let me think about small values of ( n ) to see if this makes sense.For ( n = 2 ), there's only one line, which is ( binom{2}{2} = 1 ). But the problem says "if ( n ) points are not on the same line," so for ( n = 2 ), they can't be on the same line because two points define a line. So, maybe the problem starts making sense for ( n geq 3 ).For ( n = 3 ), if the three points are not on the same line, they form a triangle, and there are three distinct lines. So, that matches the statement: at least three lines.For ( n = 4 ), if the four points are not all on the same line, how many lines do we have? If no three are on the same line, then it's ( binom{4}{2} = 6 ) lines. But if three are on the same line, then we have one line with three points and the other lines connecting the remaining points. So, how many lines would that be?If three points are on one line, say points A, B, and C, and the fourth point D is not on that line, then the lines are: AB, AC, BC (which are all the same line), AD, BD, CD. So, that's 1 line for ABC and 3 lines connecting D to A, B, and C. So, total lines are 1 + 3 = 4 lines. Which is equal to ( n = 4 ). So, that works.If four points are arranged such that two pairs are on two different lines, like a rectangle, then we have four lines: the two sides and the two diagonals. Wait, no, in a rectangle, there are four sides and two diagonals, so that's six lines. But if we have four points with two lines each containing two points, then we have two lines, but that would mean all four points are on two lines, but the problem states that not all four are on the same line, so that's allowed. But in that case, how many distinct lines do we have?If two points are on one line and the other two points are on another line, then we have two lines. But that's only two lines, which is less than ( n = 4 ). Wait, but the problem says "if ( n ) points are not on the same line," so in this case, they are on two lines, but not all on one line. So, does the problem still hold?Wait, maybe I'm misunderstanding. The problem says "if ( n ) points are not on the same line," which means that not all ( n ) points lie on a single line. It doesn't say anything about them not lying on multiple lines. So, in the case where four points are on two lines, each containing two points, then the total number of lines is two, which is less than ( n = 4 ). So, does that contradict the problem statement?Hmm, maybe I need to re-examine the problem. It says "if ( n ) points are not on the same line, then there are at least ( n ) different lines connecting them." So, in the case where four points are on two lines, each with two points, the total number of lines is two, which is less than four. So, that would mean the statement is false in that case. But the problem says it's true, so maybe I'm missing something.Wait, perhaps the problem assumes that no three points are on the same line. Because in the case where two points are on one line and two on another, that's allowed, but if three points are on a line, then the number of lines would be more.Wait, let's go back to the problem statement. It just says "if ( n ) points are not on the same line." It doesn't specify anything about three points being on the same line or not. So, in the case where four points are on two lines, each with two points, that's allowed, and the number of lines is two, which is less than four. So, that would mean the statement is false.But the problem says it's true, so maybe I'm misinterpreting the problem. Maybe it's saying that if the ( n ) points are not all on the same line, then the number of lines connecting them is at least ( n ). So, in the case where four points are on two lines, each with two points, the number of lines is two, which is less than four, so that would contradict the problem.Wait, maybe the problem assumes that no three points are on the same line. Because in that case, the number of lines would be ( binom{n}{2} ), which is more than ( n ) for ( n geq 3 ). But the problem doesn't specify that.Alternatively, maybe the problem is considering that if the points are not all on the same line, then there must be at least ( n ) lines, regardless of how they are arranged. So, even if some points are on the same line, as long as not all are on the same line, the number of lines is at least ( n ).But in the case of four points on two lines, each with two points, the number of lines is two, which is less than four. So, that would mean the statement is false. Therefore, maybe the problem assumes that no three points are on the same line.Alternatively, perhaps the problem is considering that if the points are not all on the same line, then the number of lines is at least ( n ), but in some cases, it can be more. So, maybe the minimal number of lines is ( n ), but it can be higher.Wait, let's think about it differently. Suppose we have ( n ) points, not all on the same line. What is the minimal number of lines that can connect them?If we arrange the points such that as many as possible are on the same line, but not all, then we can minimize the number of lines.For example, if ( n - 1 ) points are on one line, and the last point is off that line, then the number of lines would be ( (n - 1) ) lines connecting the last point to each of the ( n - 1 ) points on the line, plus the one line containing the ( n - 1 ) points. So, total lines would be ( (n - 1) + 1 = n ).Ah, so in this case, the number of lines is exactly ( n ). So, that's the minimal case.If we have more points not on the same line, then the number of lines would increase.So, in the case where ( n - 1 ) points are on one line and the last point is off that line, we get exactly ( n ) lines. Therefore, the minimal number of lines is ( n ), and in other configurations, it can be more.Therefore, the statement is true: if ( n ) points are not all on the same line, then there are at least ( n ) different lines connecting them.So, to summarize, the minimal number of lines occurs when ( n - 1 ) points are on one line, and the last point is off that line, resulting in ( n ) lines. Any other configuration would result in more lines.Therefore, the proof would involve showing that the minimal number of lines is ( n ), achieved when ( n - 1 ) points are colinear and the last point is not, and in all other cases, the number of lines is greater than ( n ).I think that makes sense. So, the key idea is that if you have ( n ) points not all on the same line, the minimal number of lines is ( n ), and thus, there are at least ( n ) different lines connecting them.